Container 1 has 10 items 3 of which are defective Container

Container 1 has 10 items, 3 of which are defective. Container 2 has 6 items, 2 of which are defective. If one item is drawn independently from each container find the probability distribution for X defined as the number of defective items drawn (Hint: You have to use both multiplicative and additive rules to find P(X=1), whereas P(X=0) and P(X=2) can be found only by multiplicative rules).

Solution

P(x=0) = 7/10*4/6 = 7/15
P(x=1) = 3/10 * 4/6 + 7/10 * 2/6 = 13/30
p(x=2) =1 - p(x=0) - p(x=1) = 1/10